Argument of complex number MCQ - Practice Questions & Answers

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13 Questions around this concept.

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Let w1 be the point obtained by the rotation of z1 = 5 + 4i about the origin through a right angle in the
anticlockwise direction, and w2 be the point obtained by the rotation of z2 = 3 + 5i about the origin through a right angle in the clockwise direction. Then the principal argument of w1 – w2 is equal to :

$\pi-\tan ^{-1} \frac{8}{9}$

$-\pi+\tan ^{-1} \frac{8}{9}$

$\pi-\tan ^{-1} \frac{33}{5}$

$-\pi+\tan ^{-1} \frac{33}{5}$

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Argument of complex number

If a complex number z = x + iy is represented by a point P in Argand plane and OP forms some angle with positive x-axis, let's denote it with ?, then ? is called the argument of z.

$\\\mathrm{\tan\theta=\frac{PM}{OM}}\\\mathrm{\tan\theta=\frac{y}{x}=\frac{Im(z)}{Re(z)}\Rightarrow \theta=\tan^{-1}\frac{y}{x}}\\\mathrm{\arg(z)=\theta=\tan^{-1}\frac{y}{x}}$

If ? lies between -? < ? ≤ ?, then ? is called principal argument. Value of argument differs depending on which quadrant point (x,y) lies.

If it lies in 1^st quadrant then it is ? (acute angle)

If the point lies in 2^nd quadrant, then $arg(z)=\theta =\pi - \tan ^{-1}\frac{y}{|x|}$

So it will be an obtuse +ve angle

If the point lies in lies in 3^rd quadrant then $arg(z)=\theta =-\pi + \tan ^{-1}\frac{y}{x}$

It will be an obtuse -ve angle

If the point lies in 4^th quadrant then $arg(z)=\theta =- \tan ^{-1}\frac{|y|}{x}$

It will be -ve acute angle

Note:

$\\\mathrm{If\;\arg(z)=\frac{\pi}{2}\;\;or\;-\frac{\pi}{2},\;z\;is\;purely\;imaginary.}\\\mathrm{If\;\arg(z)=0\;\;or\;\pi,\;z\;is\;purely\;real.}$

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